Powers (how #2^(2017/2)=sqrt2*2^1008# works)?
I have #(1-i)^2017=?#
I know that #(cistheta)^n=cis(ntheta)#
#=> ... =>#
#=2^1008-i2^1008#
The problem is HOW am I solving:
#2^(2017/2)=sqrt2*2^1008#
I have
I know that
The problem is HOW am I solving:
1 Answer
Nov 29, 2017
First remember that:
We know that
By our second and third rule, we know that
When simplified, it becomes